After analysis it is obvious that each segment, in each sector, is exactly 1/2 of the sector width. So if we traverse each concentric circle, there is no distinction between segments. If we use a RLE (Run Length Encoding) Algorithm we can prove that each segment encodes to the same value and would therefore contain the 'same' data. Therefore, any circular traversal of the pattern would fail to reveal any data. If on the other hand we look at each radial ray (from the centre), we could travel up this ray and ascertain that each segment section is either left or right of the centre line. This is a variant of Huffman Encoding. Then we would have to know what encodings have been designated for what Number/Letter/Symbol. I did that Huffman Tree coding and have now got a program that will encode any message in a crop circle. The problem is that it doesn't look anything like that one! So I got to thinking, what if, I am intellectualising it too much? What if, I was looking too deeply for the answer. So I tried a different approach. Standing at the centre of the circle: If we assume left = 0 right = 1: 1: l,l,r,l,r,l,l,r = 0,0,1,0,1,0,0,1 = 0010 1001 * 2: l,l,r,r,l,l,l,r = 0,0,1,1,0,0,0,1 = 0011 0001 3: l,l,r,r,r,r,l,r = 0,0,1,1,1,1,0,1 = 0011 1101 4: l,l,r,r,l,l,l,l = 0,0,1,1,0,0,0,0 = 0011 0000 5: l,r,r,l,l,r,l,r = 0,1,1,0,0,1,0,1 = 0110 0101 6: l,r,l,r,r,r,r,l = 0,1,0,1,1,1,1,0 = 0101 1110 7: l,l,r,l,r,l,l,l = 0,0,1,0,1,0,0,0 = 0010 1000 8: l,r,r,l,r,l,l,l = 0,1,1,0,1,0,0,0 = 0110 1000 9: l,r,r,l,r,l,l,r = 0,1,1,0,1,0,0,1 = 0110 1001 ~ 10: l,l,r,l,r,l,l,r = 0,0,1,0,1,0,0,1 = 0010 1001 * 11: l,r,r,r,l,l,l,l = 0,1,1,1,0,0,0,0 = 0111 0000 12: l,r,r,l,r,l,l,r = 0,1,1,0,1,0,0,1 = 0110 1001 ~ We have two similiar rays (1,10) & (9,12). We can see that there are eight concentric rings so we have 8 bits of information in each ray. Using 8bit ACII we would have 1: 0010 1001 = ) 2: 0011 0001 = 1 3: 0011 1101 = = 4: 0011 0000 = 0 5: 0110 0101 = e 6: 0101 1110 = ^ 7: 0010 1000 = ( 8: 0110 1000 = h 9: 0110 1001 = i 10: 0010 1001 = ) 11: 0111 0000 = p 12: 0110 1001 = i Which says: )1=0e^(hi)pi (arrange them how you like) Which says [ ) | 1 = e^0 | (hi) | pi ] ===> smiley, 1=e^0, hi, pi NOTE: I designated the left tree, 0 (Zero) and the right tree 1 (One), Why? Well if we did the opposite, we would have for say [ ray1 |-----> 1101 0110 ] and so on for all of the codes. There being a 11 in the first two position makes this unlikely as any encoding would have to start from the lowest value 00. We could look deeper, but I can't really see any point. I was hoping for something more dramatic, but alas it is just saying hi, smiling at us, telling us that e^0=1, and stating pi. Steven Murray 27/05/2010 |